It's not clear from your question you're trying to do. A DS12887 "chip" has its own internal battery. It's actually not a chip, but a module containing a DS12885 chip, a crystal and a small lithium battery. The module only has one power supply connection, a +5V VCC pin and this pin must be connected to the +5V power supply during normal operation. It has no means of connecting an external battery. It does not use a CR2032 battery.
If you're saying that your motherboard has a CR2032 socket and you want connect +5V to it instead of replacing the battery with fresh a CR2032 then you shouldn't need to do anything. Like Cody Gray said, just remove the dead CR2032 battery and your motherboard should boot.
If you're saying you've broken open a DS12887 module and are attempting to replace the internal battery then you probably don't need to do anything except remove the battery for the motherboard to boot. Note that this battery is not a CR2032, it's a much smaller battery like a CR1232. The DS12885 chip inside the module just needs to be powered with +5V through its VCC pin, and this is required for it to be fully functional. The motherboard must already be supplying +5V to the DS12887 otherwise it would have never have worked.
You can confirm all this with the DS12887's datasheet. For example for the VBAT connection it says:
Connection for a Primary Battery. (DS12885 Only.) Battery voltage must be held
between the minimum and maximum limits for proper operation. If a backup
supply is not supplied, VBAT must be grounded. Connect the battery directly to
the VBAT pin. Diodes in series between the VBAT pin and the battery may
prevent proper operation.
Notice that only the DS12885 has a VBAT pin, and that it doesn't actually need to be connected to a battery. If you've cracked open a DS12887 you might need to ground the DS12885 chip's VBAT pin, but you don't need to supply it any power. Supplying +5V to battery terminals inside the module risks damaging it, supplying it a dropped down 2.5V - 4V from the power supply won't accomplish anything.
The description for the VCC pin reads:
When VCC is applied within normal limits, the device is fully accessible
and data can be written and read. When VCC is below VPF reads and writes
are inhibited.
Notice how the chip cannot be accessed unless its being supplied +5V through the VCC pin. This means when the computer is powered on the chip must be supplied with +5V for it to work, so this must already be the case.
If you're not sure whether you actually have DS12887 module then according to App Note 503: Replacing the DS1287/DS12887
Real-Time Clock in a Personal Computer here's how you can tell:
If your computer contains a Nickel Cadmium or a coin cell battery attached to the motherboard, or
uses a battery connected to the motherboard via a wire harness, then your computer does not use a
clock module. If your PC does use a Dallas clock module, it will look something like the
following:
Finally, if you are in fact ultimately trying to replace the internal battery in a DS12887 then you should consider simply replacing the entire module instead. They're still being made and can be bought for $10 or less online.
